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Source. S. Fan, Strongly complete sets and a conjecture of Erdős, arXiv:2607.14071v5 (16 September 2026); Remark 4.2 on p. 20 (the second paragraph of Remark 4.1 of v4, p. 19, with the same content), with the definition (1.5) on p. 3, the definitions (1.8) and (1.9) and Corollary 1.2 on p. 4 and the discussion of Hegyvári's conjecture on p. 4. The artifacts are identified on the source card.

Read depth. Claims checked: the remark, the definitions and Corollary 1.2 were read clause by clause in the text layer of v5 and compared with v4; the remark's half-page argument was read through and not independently reviewed; Corollary 1.2 rests on Theorem 1.1, whose proof (Section 4) was not read. A preprint.

Statement

Definitions (pp. 3--4). For $A\subseteq\mathbb N$, (1.5) (p. 3) is $\sum_{a\in A}\|a\theta\|=\infty$ for every $\theta\in\mathbb T\setminus\{0\}$ (equivalently, the paper's spectrum $H_1(A)$ is $\{0\}$). $M_\rho^*$ (1.8) is the smallest positive integer $M$ with this property: whenever $A$ satisfies (1.5) and $|A\cap(\rho^k,\rho^{k+1}]|\ge M$ for all large enough $k$, $A$ is strongly complete. For $\alpha,\beta>0$,

$$A_{\alpha,\beta}=\{\lfloor2^k\alpha\rfloor,\lfloor2^k\beta\rfloor:k\ge0\}\setminus\{0\} \qquad(1.9);$$

$\alpha\sim\beta$ means $\alpha/\beta=2^n$ for some $n\in\mathbb Z$, and $\alpha$ is a dyadic rational when $\alpha\sim n$ for some nonzero integer $n$. Hegyvári's conjecture, as the paper reports it (p. 4): if $\alpha\not\sim\beta$ and $\alpha,\beta$ are not both dyadic rationals, then $A_{\alpha,\beta}$ is complete. Corollary 1.2 (p. 4). If $A\subseteq\mathbb N$ satisfies (1.5) and has at least five elements in $(2^k,2^{k+1}]$ for all large enough $k$, then $A$ is strongly complete; that is, $M_2^*\le5$. Remark 4.2 (p. 20). If $M_2^*$ were $2$, then $A_{\alpha,\beta}$ would be strongly complete, and so complete, for all $\alpha,\beta>0$ with $\alpha\not\sim\beta$ that are not both dyadic rationals; Hegyvári's conjecture would follow.

Proof pointer

Remark 4.2 (p. 20). Label the parameters so that $\alpha$ is not a dyadic rational, and write $U_K(x)=\{\lfloor2^kx\rfloor:k\ge K\}$. Because $\alpha\not\sim\beta$, the rays $U_0(\alpha)$ and $U_0(\beta)$ meet in a finite set. Multiplying $\alpha$ and $\beta$ by suitable powers of $2$ moves both into $(1/2,1]$, where they differ (as $\alpha\not\sim\beta$) and $\alpha$ is still not a dyadic rational; this changes $A_{\alpha,\beta}$ by finitely many elements, which affects neither (1.5) nor strong completeness. Then, for every large enough $k_0$, $A_{\alpha,\beta}$ is the disjoint union of $U_{k_0}(\alpha)$, $U_{k_0}(\beta)$ and a finite set $B$, and for each $k\ge k_0$ the integers $\lfloor2^{k+1}\alpha\rfloor$ and $\lfloor2^{k+1}\beta\rfloor$ differ and both lie in $(2^k,2^{k+1}]$, so each such interval holds at least two elements of $A_{\alpha,\beta}$. Since $\alpha$ is not a dyadic rational, $\lfloor2^{k+1}\alpha\rfloor=2\lfloor2^k\alpha\rfloor+1$ for infinitely many $k$; for each such $k$ and each $\theta\in\mathbb T\setminus\{0\}$ the triangle inequality gives $\|\theta\|\le\|\lfloor2^{k+1}\alpha\rfloor\theta\|+2\|\lfloor2^k\alpha\rfloor\theta\|$, so the sum in (1.5) diverges. With two elements in every such interval and (1.5), $M_2^*=2$ would make $A_{\alpha,\beta}$ strongly complete. Read through; not reviewed.

Dependencies

Theorem 1.1 and Corollary 1.2 of the same paper for $M_2^*\le5$; v5's Remark 4.1 for $M_2^*\ge2$; Hegyvári's 1989 paper for the conjecture and its proved case (cited as the paper's [25]; not held).

Bears on

• Problem 354: context. The remark's hypothesis $M_2^*=2$ is unproved ($2\le M_2^*\le5$ is what the paper gives), so it resolves neither the irrational-ratio question, which the site-accepted Lean proof on the Conjectures.io card answers, nor the rational-ratio cases of Hegyvári's conjecture with neither number a dyadic rational (the paper reports that Hegyvári confirmed the case where exactly one is, p. 4), which remain open; the bounty site's review cites the paper (v4) as leaving "the relevant two-ray case unresolved".